这是我在单击添加按钮时在表上添加动态行的代码。现在我需要将输入的值输入文本框并提交给DB。
<div id="addinput">
<p>
<button name="add" onclick="myFunction(this)" value="'.$btn_count.'"><font color="#FF0000" size="4">+</font></button>
</p>
</div>
</td>
这是我向桌面添加动态行的功能,就在我们点击
的行的下方<script>
function myFunction(y)
{
var f = y.parentNode;
var d = f.parentNode;
var g = d.parentNode;
var c =g.parentNode.rowIndex;
var table = document.getElementById("myTable");
var row = table.insertRow(c+1);
var cell1 = row.insertCell(0);
var cell2 = row.insertCell(1);
var cell3 = row.insertCell(2);
var cell4 = row.insertCell(3);
var cell5 = row.insertCell(4);
var cell6 = row.insertCell(5);
var cell7 = row.insertCell(6);
cell1.innerHTML = "<form name='frm_sub' id='frm_id' action='#'><input type=text name='txt_id' id='id_id' size=10 placeholder=Id>";
cell2.innerHTML = "<input type=text name='txt_item' id='id_item' size=10 placeholder=Item>";
cell3.innerHTML = "<input type=text name='txt_cost' id='id_cost' size=10 placeholder=0.00>";
cell4.innerHTML = "<input type=text name='txt_wp' id='id_wp' size=10 placeholder=0.00>";
cell5.innerHTML = "<input type=text name='txt_rp' id='id_rp' size=10 placeholder=0.00>";
cell6.innerHTML = "<input type='button' name='submit' id='id_submitBtn' value='Save' size=10 onclick='form_submit()' ></form>";
document.getElementById("myTable");
}
function form_submit()
{
var item_name = $("input#id_item").val();
var cost = $("input#id_cost").val();
var wp = $("input#id_wp").val();
var rp = $("input#id_rp").val();
var dataString = 'item_name='+ item_name + '&cost=' + cost + '&wp=' + wp+ '&rp=' + rp;
alert(dataString);
$.ajax({
type: "POST",
url:newAdd.php,
dataType: "json",
type: "POST",
data: {
items : $("#id_item").val(),
cost : $("#id_cost").val(),
wp : $("#id_wp").val(),
rp : $("#id_rp").val(),
},
success: function () {
alert("Success");
}
});
}
在newAdd.php我写了
echo$itemName =$_POST["txt_item"];
$insert = mysql_query('INSERT INTO item (item) VALUES ('$itemName');
但这不起作用..
答案 0 :(得分:4)
if you are getting that alert then this will work
function form_submit()
{
var item_name = $("input#id_item").val();
var cost = $("input#id_cost").val();
var wp = $("input#id_wp").val();
var rp = $("input#id_rp").val();
var dataString = 'item_name='+ item_name + '&cost=' + cost + '&wp=' + wp+ '&rp=' + rp;
$.ajax({
type: "post",
url: "newAdd.php",
data:dataString,
success: function(msg){
alert('success');
}
});
}
答案 1 :(得分:0)
试试这个ajax请求。希望这会对你有所帮助 -
$.ajax({
type: "POST",
url:'newAdd.php',
dataType: "json",
data: {
items : $("#id_item").val(),
cost : $("#id_cost").val(),
wp : $("#id_wp").val(),
rp : $("#id_rp").val()
},
success: function () {
alert("Success");
}
});